3.43 \(\int \frac {\log (c (a+\frac {b}{x^2})^p)}{x^3} \, dx\)

Optimal. Leaf size=35 \[ \frac {p}{2 x^2}-\frac {\left (a+\frac {b}{x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{2 b} \]

[Out]

1/2*p/x^2-1/2*(a+b/x^2)*ln(c*(a+b/x^2)^p)/b

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Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2454, 2389, 2295} \[ \frac {p}{2 x^2}-\frac {\left (a+\frac {b}{x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x^2)^p]/x^3,x]

[Out]

p/(2*x^2) - ((a + b/x^2)*Log[c*(a + b/x^2)^p])/(2*b)

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps

\begin {align*} \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x^3} \, dx &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \log \left (c (a+b x)^p\right ) \, dx,x,\frac {1}{x^2}\right )\right )\\ &=-\frac {\operatorname {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,a+\frac {b}{x^2}\right )}{2 b}\\ &=\frac {p}{2 x^2}-\frac {\left (a+\frac {b}{x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 0.97 \[ \frac {1}{2} \left (\frac {p}{x^2}-\frac {\left (a+\frac {b}{x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{b}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x^2)^p]/x^3,x]

[Out]

(p/x^2 - ((a + b/x^2)*Log[c*(a + b/x^2)^p])/b)/2

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fricas [A]  time = 0.43, size = 41, normalized size = 1.17 \[ \frac {b p - b \log \relax (c) - {\left (a p x^{2} + b p\right )} \log \left (\frac {a x^{2} + b}{x^{2}}\right )}{2 \, b x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x^2)^p)/x^3,x, algorithm="fricas")

[Out]

1/2*(b*p - b*log(c) - (a*p*x^2 + b*p)*log((a*x^2 + b)/x^2))/(b*x^2)

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giac [B]  time = 0.17, size = 65, normalized size = 1.86 \[ -\frac {a p \log \left (a x^{2} + b\right )}{2 \, b} + \frac {a p \log \relax (x)}{b} - \frac {p \log \left (a x^{2} + b\right )}{2 \, x^{2}} + \frac {p \log \left (x^{2}\right )}{2 \, x^{2}} + \frac {b p - b \log \relax (c)}{2 \, b x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x^2)^p)/x^3,x, algorithm="giac")

[Out]

-1/2*a*p*log(a*x^2 + b)/b + a*p*log(x)/b - 1/2*p*log(a*x^2 + b)/x^2 + 1/2*p*log(x^2)/x^2 + 1/2*(b*p - b*log(c)
)/(b*x^2)

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maple [A]  time = 0.05, size = 50, normalized size = 1.43 \[ \frac {a p}{2 b}-\frac {a \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{2 b}+\frac {p}{2 x^{2}}-\frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(a+b/x^2)^p)/x^3,x)

[Out]

-1/2/b*ln(c*(a+b/x^2)^p)*a-1/2/x^2*ln(c*(a+b/x^2)^p)+1/2*a/b*p+1/2*p/x^2

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maxima [A]  time = 0.65, size = 54, normalized size = 1.54 \[ -\frac {1}{2} \, b p {\left (\frac {a \log \left (a x^{2} + b\right )}{b^{2}} - \frac {a \log \left (x^{2}\right )}{b^{2}} - \frac {1}{b x^{2}}\right )} - \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x^2)^p)/x^3,x, algorithm="maxima")

[Out]

-1/2*b*p*(a*log(a*x^2 + b)/b^2 - a*log(x^2)/b^2 - 1/(b*x^2)) - 1/2*log((a + b/x^2)^p*c)/x^2

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mupad [B]  time = 0.24, size = 47, normalized size = 1.34 \[ \frac {p}{2\,x^2}-\frac {\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{2\,x^2}-\frac {a\,p\,\ln \left (a\,x^2+b\right )}{2\,b}+\frac {a\,p\,\ln \relax (x)}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b/x^2)^p)/x^3,x)

[Out]

p/(2*x^2) - log(c*(a + b/x^2)^p)/(2*x^2) - (a*p*log(b + a*x^2))/(2*b) + (a*p*log(x))/b

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sympy [A]  time = 5.65, size = 58, normalized size = 1.66 \[ \begin {cases} - \frac {a p \log {\left (a + \frac {b}{x^{2}} \right )}}{2 b} - \frac {p \log {\left (a + \frac {b}{x^{2}} \right )}}{2 x^{2}} + \frac {p}{2 x^{2}} - \frac {\log {\relax (c )}}{2 x^{2}} & \text {for}\: b \neq 0 \\- \frac {\log {\left (a^{p} c \right )}}{2 x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(a+b/x**2)**p)/x**3,x)

[Out]

Piecewise((-a*p*log(a + b/x**2)/(2*b) - p*log(a + b/x**2)/(2*x**2) + p/(2*x**2) - log(c)/(2*x**2), Ne(b, 0)),
(-log(a**p*c)/(2*x**2), True))

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